Answer
$\frac{dr}{d\theta}=\frac{{-2\sin\theta}}{({1-\cos\theta})^2}$
Work Step by Step
$r=(\frac{\sin\theta}{\cos\theta-1})^2$
$r=\frac{\sin^2\theta}{(\cos\theta-1)^2}=\frac{1-\cos^2\theta}{(1-\cos\theta)^2}$
$r=\frac{1+\cos\theta}{1-\cos\theta}$
0n differentiating the above
$\frac{dr}{d\theta}=\frac{d(\frac{1+\cos\theta}{1-\cos\theta})}{d\theta}$
$\frac{dr}{d\theta}=\frac{{(1-\cos\theta)\frac{d({1+\cos\theta})}{d\theta}}-({1+\cos\theta})\frac{d({1-\cos\theta})}{d\theta}}{({1-\cos\theta})^2}$
$\frac{dr}{d\theta}=\frac{{(1-\cos\theta){({-\sin\theta})}}-({1+\cos\theta})({\sin\theta})}{({1-\cos\theta})^2}$
$\frac{dr}{d\theta}=\frac{{-2\sin\theta}}{({1-\cos\theta})^2}$
